182k views
4 votes
Graham and Hunter are circus performers. A cable lifts Graham into the air at a constant speed of 1.5 ft/s. When Graham’s arms are 18 ft above the ground, Hunter, who is standing directly underneath Graham, throws Graham a ball as the cable continues to lift him higher. Hunter throws the ball from a position 5 ft above the ground with an initial velocity of 24 ft/s. Which system of equations can be used to model this situation?

A) h=18+1.5t
h=5+24t-16t^2

B) h=18+1.5t
h=5+24t+16t^2

C) h=18+1.5t
h=5+24t

D) h=18t+1.5t-16t^2
h=5+24t-16t^2

User Sam Alex
by
6.7k points

2 Answers

5 votes

Answer:

A

Explanation:

EDGE 2020

User Afridi
by
6.9k points
5 votes

Answer: Hello there!

this type of equations in one dimension (when all the factors are constants) are written as:

h = initial position + initial velocity*t + (acceleration/2)*t^2

First, let's describe the hunter's equation:

We know that Graham moves with a velocity of 1.5 ft/s, and when he is 18 ft above the ground, Hunter throws the ball, and because Graham is pulled with a cable, he is not affected by gravity.

If we define t= 0 when Graham is 18 ft above the ground, the equation for Graham height (in feet) is:

h = 18 + 1.5t

where t in seconds.

Now, the equation for the ball:

We know that at t= 0, the ball is thrown from an initial distance of 5ft, with an initial velocity of 24ft/s and is affected by gravity acceleration g, where g is equal to: 32.2 ft/s (notice that the gravity pulls the ball downwards, so it will have a negative sign)

the equation for the ball is:

h = 5 + 24t - (32.2/2)t^2 = 5 + 24t - 16.1t^2

So the system is:

h = 18 + 1.5t

h = 5 +24t - 16.1t^2

so the right answer is A

User Agirardello
by
6.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.